I asked this question, Calculating the expression for the derivative of a Gaussian process, some time ago, and now I am interested in an extension to the question. So originally I wanted to know the following: Given $$f(x_1),f(x_2),....,f(x_n)\sim GP(X\beta,\sigma^2R)$$ where $R$ is the Gaussian correlation function


then the distribution of $f'(x_1),f'(x_2),...,f'(x_n)$ is $$ f'(x_j) \sim \text{GP}\left( \beta_j, \sigma^2 \frac{2}{\phi_j} R\right) $$

Now what I would ultimately like to know is whether or not it is possible to calculate the posterior predictive process of the derivative process based on NON DERIVATIVE observations, i.e., is the following quantity defined?

$$f'(x_1),f'(x_2),...,f'(x_n)\,|\,\,f(x_1),f(x_2),...,f(x_n)\sim ???$$

@g g seems to suggest it can be done here: Derivative of a Gaussian Process but I don't follow the argument.

  • $\begingroup$ What is $f$? Usually if we know something distribution of $Y=f(X)$ it is hard to say something about distribution of $f'(X)$ if we do not know something specific about $g$ in $f'(x)=g(f(x))$. $\endgroup$
    – mpiktas
    Apr 15, 2016 at 13:23
  • $\begingroup$ I think this is a duplicate of the question linked in the last sentence (and the bounty should be put into the linked question to encourage more clear answers there) but apparently voting to close is impossible when there is an open bounty. In any case, the title here is somewhat misleading as this is about Gaussian processes specifically, not stochastic processes in general. $\endgroup$ Apr 15, 2016 at 13:26
  • $\begingroup$ @mpiktas $f$ is a Gaussian process (note that $x$ is not intended to be random, I'm not sure if your use of capital $X$ implies you misunderstood the question as being about evaluating some fixed function at a random $x$) $\endgroup$ Apr 15, 2016 at 13:28
  • $\begingroup$ @JuhoKokkala, from what I recall the Gaussian process is a stochastic process, whose fdds are Gaussian. And as far as I remember the derivative is not defined for such type of mathematical object. $\endgroup$
    – mpiktas
    Apr 15, 2016 at 13:33
  • $\begingroup$ @mpiktas, For example Rasmussen and Williams (see Eq. 4.4, gaussianprocess.org/gpml/chapters/RW4.pdf)) define the derivative in the mean square sense (this exists for some Gaussian processes, depending on the covariance function). I have no idea whether this implies existence of the derivative in some other sense (e.g. is there a version of the process for which the realizations are almost surely pathwise differentiable), but the present question seems valid in any case. See also discussion in the linked question. $\endgroup$ Apr 15, 2016 at 13:43

1 Answer 1


I will consider the case of zero mean random process for simplicity. The generalization for nonzero mean is straightforward.

We need two facts from random processes theory:

  1. Covariance between random process and its' derivative is $$ \mathrm{cov}\left(f(x), \frac{\partial f(x')}{\partial x_i}\right) = \frac{\partial R(x, x')}{\partial x'_i}.$$
  2. Joint distribution of derivatives and values of Gaussian processes is multivariate normal.

The first fact allows you to calculate covariance matrix for joint of derivatives and values of a realization of Gaussian process $K$. Second fact allows you to get formula for marginal distribution of derivatives given values. Formula is similar to usual formula for Gaussian process $\hat{f}'(x) = k K^{-1} f$, but you need to insert specific covariance calculated with the formula given above.


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